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Dipole Moments and Polar Molecules 10.2 H F electron rich region electron poor region    = Q x r Q is the charge r is the distance between charges.

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Presentation on theme: "Dipole Moments and Polar Molecules 10.2 H F electron rich region electron poor region    = Q x r Q is the charge r is the distance between charges."— Presentation transcript:

1 Dipole Moments and Polar Molecules 10.2 H F electron rich region electron poor region    = Q x r Q is the charge r is the distance between charges 1 D = 3.36 x 10 -30 C m D= debye unit

2 10.2

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4 Which of the following molecules have a dipole moment? H 2 O, CO 2, SO 2, and CH 4 O H H dipole moment polar molecule S O O CO O no dipole moment nonpolar molecule dipole moment polar molecule C H H HH no dipole moment nonpolar molecule

5 Does CH 2 Cl 2 have a dipole moment? 10.2

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7 Chemistry In Action: Microwave Ovens

8 Bond Dissociation EnergyBond Length H2H2 F2F2 436.4 kJ/mole 150.6 kJ/mole 74 pm 142 pm Valence bond theory – bonds are formed by sharing of e - from overlapping atomic orbitals. Overlap Of 2 1s 2 2p How does Lewis theory explain the bonds in H 2 and F 2 ? Sharing of two electrons between the two atoms. 10.3

9 10.4

10 Change in electron density as two hydrogen atoms approach each other. 10.3

11 Valence Bond Theory and NH 3 N – 1s 2 2s 2 2p 3 3 H – 1s 1 If the bonds form from overlap of 3 2p orbitals on nitrogen with the 1s orbital on each hydrogen atom, what would the molecular geometry of NH 3 be? If use the 3 2p orbitals predict 90 0 Actual H-N-H bond angle is 107.3 0 10.4

12 Hybridization – mixing of two or more atomic orbitals to form a new set of hybrid orbitals. 1.Mix at least 2 nonequivalent atomic orbitals (e.g. s and p). Hybrid orbitals have very different shape from original atomic orbitals. 2.Number of hybrid orbitals is equal to number of pure atomic orbitals used in the hybridization process. 3.Covalent bonds are formed by: a.Overlap of hybrid orbitals with atomic orbitals b.Overlap of hybrid orbitals with other hybrid orbitals 10.4

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15 Predict correct bond angle

16 Formation of sp Hybrid Orbitals 10.4

17 Formation of sp 2 Hybrid Orbitals 10.4

18 # of Lone Pairs + # of Bonded Atoms HybridizationExamples 2 3 4 5 6 sp sp 2 sp 3 sp 3 d sp 3 d 2 BeCl 2 BF 3 CH 4, NH 3, H 2 O PCl 5 SF 6 How do I predict the hybridization of the central atom? Count the number of lone pairs AND the number of atoms bonded to the central atom 10.4

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20 10.5

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22 Sigma bond (  ) – electron density between the 2 atoms Pi bond (  ) – electron density above and below plane of nuclei of the bonding atoms 10.5

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26 Sigma (  ) and Pi Bonds (  ) Single bond 1 sigma bond Double bond 1 sigma bond and 1 pi bond; sp 2 hybridized 2 Double or 1 Triple bond 1 sigma bond and 2 pi bonds; sp hybridized How many  and  bonds are in the acetic acid (vinegar) molecule CH 3 COOH? C H H CH O OH  bonds = 6 + 1 = 7  bonds = 1 10.5


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